2. NUMERICAL DIFFERENTIATION
The derivative of f (x) at x0 is:
f ′( x 0 ) =
Limit
h→ 0
f ( x0 + h) − f ( x0 )
h
An approximation to this is:
f ( x0 + h) − f ( x0 )
f ′( x 0 ) ≈
h
for small values of h.
Forward Difference
Formula
3. Let f ( x ) = ln x and x0 = 1.8
Find an approximate value for f ′(1.8)
h
f (1.8)
f (1.8 + h)
0.1 0.5877867 0.6418539
f (1.8 + h) − f (1.8)
h
0.5406720
0.01 0.5877867 0.5933268
0.5540100
0.001 0.5877867 0.5883421
0.5554000
The exact value of
f ′(1.8) = 0.55 5
4. Assume that a function goes through three points:
( x0 , f ( x0 ) ) , ( x1 , f ( x1 ) ) and ( x2 , f ( x2 ) ) .
f ( x) ≈ P( x)
P ( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x2 )
Lagrange Interpolating Polynomial
5. P ( x ) = L0 ( x ) f ( x0 ) + L1 ( x ) f ( x1 ) + L2 ( x ) f ( x 2 )
( x − x1 )( x − x2 )
P( x) =
f ( x0 )
( x0 − x1 )( x0 − x2 )
( x − x0 )( x − x2 )
+
f ( x1 )
( x1 − x0 )( x1 − x2 )
( x − x0 )( x − x1 )
+
f ( x2 )
( x2 − x0 )( x2 − x1 )
6. f ′( x ) ≈ P ′( x )
2 x − x1 − x2
P ′( x ) =
f ( x0 )
( x0 − x1 )( x0 − x2 )
2 x − x0 − x2
+
f ( x1 )
( x1 − x0 )( x1 − x2 )
2 x − x0 − x1
+
f ( x2 )
( x2 − x0 )( x2 − x1 )
8. − 3h
− 2h
−h
P ′( x 0 ) =
f ( x0 ) +
f ( x1 ) + 2 f ( x 2 )
2
2
2h
−h
2h
1
{ − 3 f ( x0 ) + 4 f ( x1 ) − f ( x2 ) }
P ′( x0 ) =
2h
Three-point formula:
1
{ − 3 f ( x0 ) + 4 f ( x0 + h) − f ( x0 + 2h) }
f ′( x0 ) ≈
2h
9. If the points are equally spaced with x0 in the middle:
x1 = x0 − h and x 2 = x0 + h
2 x0 − ( x0 − h) − ( x0 + h)
P ′( x0 ) =
f ( x0 )
{ x0 − ( x0 − h)}{ ( x0 − ( x0 + h)}
2 x0 − x0 − ( x0 + h)
+
f ( x1 )
{ ( x0 − h) − x0 }{ ( x0 − h) − ( x0 + h)}
2 x0 − x0 − ( x0 − h)
+
f ( x2 )
{ ( x0 + h) − x0 }{ ( x0 + h) − ( x0 − h)}
10. 0
−h
h
P ′( x 0 ) =
f ( x0 ) + 2 f ( x1 ) + 2 f ( x 2 )
2
−h
2h
2h
Another Three-point formula:
1
{ f ( x0 + h) − f ( x0 − h) }
f ′( x0 ) ≈
2h
11. Alternate approach (Error estimate)
Take Taylor series expansion of f(x+h) about x:
2
3
h ( 2)
h ( 3)
f ( x + h) = f ( x ) + hf ′( x ) +
f ( x) +
f ( x) +
2
3!
h2 ( 2 )
h3 ( 3 )
f ( x + h) − f ( x ) = hf ′( x ) +
f ( x) +
f ( x) +
2
3!
f ( x + h) − f ( x )
h ( 2)
h2 ( 3 )
= f ′( x ) + f ( x ) +
f ( x) +
h
2
3!
.............. (1)
12. f ( x + h) − f ( x )
= f ′( x ) + O ( h)
h
f ( x + h) − f ( x )
f ′( x ) =
− O ( h)
h
f ( x + h) − f ( x )
f ′( x ) ≈
h
Forward Difference
Formula
h ( 2)
h2 ( 3 )
O ( h) = f ( x ) +
f ( x) +
2
3!
13. 4h 2 ( 2 )
8h 3 ( 3 )
f ( x + 2h) = f ( x ) + 2hf ′( x ) +
f ( x) +
f ( x) +
2
3!
4h 2 ( 2 )
8h 3 ( 3 )
f ( x + 2h) − f ( x ) = 2hf ′( x ) +
f ( x) +
f ( x) +
2
3!
f ( x + 2h ) − f ( x )
2h ( 2 )
4h 2 ( 3 )
= f ′( x ) +
f ( x) +
f ( x) +
2h
2
3!
................. (2)
14. f ( x + h) − f ( x )
h ( 2)
h ( 3)
= f ′( x ) + f ( x ) +
f ( x) +
h
2
3!
.............. (1)
2
f ( x + 2h ) − f ( x )
2h ( 2 )
4h 2 ( 3 )
= f ′( x ) +
f ( x) +
f ( x) +
2h
2
3!
................. (2)
2 X Eqn. (1) – Eqn. (2)
15. f ( x + h) − f ( x ) f ( x + 2h) − f ( x )
2
−
h
2h
2 h2 ( 3 )
6 h3 ( 4 )
= f ′( x ) −
f ( x) −
f ( x) −
3!
4!
− f ( x + 2 h) + 4 f ( x + h) − 3 f ( x )
2h
2
3
2h ( 3 )
6h ( 4 )
′( x ) −
= f
f ( x) −
f ( x) −
3!
4!
= f ′ ( x ) + O h2
( )
16. − f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x )
= f ′( x ) + O h2
2h
( )
− f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x )
f ′( x ) =
− O h2
2h
( )
− f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x )
f ′( x ) ≈
2h
Three-point Formula
( )
2 h2 ( 3 )
6 h3 ( 4 )
O h2 = −
f ( x) −
f ( x) −
3!
4!
17. The Second Three-point Formula
Take Taylor series expansion of f(x+h) about x:
h2 ( 2 )
h3 ( 3 )
f ( x + h) = f ( x ) + hf ′( x ) +
f ( x) +
f ( x) +
2
3!
Take Taylor series expansion of f(x-h) about x:
2
3
h ( 2)
h ( 3)
f ( x − h) = f ( x ) − hf ′( x ) +
f ( x) −
f ( x) +
2
3!
Subtract one expression from another
2h 3 ( 3 )
2h 6 ( 6 )
f ( x + h) − f ( x − h) = 2hf ′( x ) +
f ( x) +
f ( x) +
3!
6!
18. 2h 3 ( 3 )
2h 6 ( 6 )
f ( x + h) − f ( x − h) = 2hf ′( x ) +
f ( x) +
f ( x) +
3!
6!
f ( x + h) − f ( x − h)
h2 ( 3)
h5 ( 6 )
= f ′( x ) +
f ( x) +
f ( x) +
2h
3!
6!
f ( x + h) − f ( x − h) h 2 ( 3 )
h5 ( 6 )
f ′( x ) =
−
f ( x) −
f ( x) −
2h
3!
6!
19. f ′( x ) =
f ( x + h ) − f ( x − h)
+ O h2
2h
( )
( )
h2 ( 3)
h5 ( 6 )
O h2 = −
f ( x) −
f ( x) −
3!
6!
f ( x + h) − f ( x − h )
f ′( x ) ≈
2h
Second Three-point Formula
20. Summary of Errors
f ( x + h) − f ( x )
f ′( x ) ≈
h
Error term
Forward Difference
Formula
h ( 2)
h2 ( 3 )
O ( h) = f ( x ) +
f ( x) +
2
3!
21. Summary of Errors continued
First Three-point Formula
− f ( x + 2h ) + 4 f ( x + h ) − 3 f ( x )
f ′( x ) ≈
2h
( )
2 h2 ( 3 )
6 h3 ( 4 )
Error term O h2 = −
f ( x) −
f ( x) −
3!
4!
22. Summary of Errors continued
Second Three-point Formula
f ( x + h) − f ( x − h )
f ′( x ) ≈
2h
( )
h2 ( 3)
h5 ( 6 )
O h2 = −
f ( x) −
f ( x) −
Error term
3!
6!
23. Example:
f ( x ) = xe
x
Find the approximate value of
x
f ( x)
1.9
12.703199
2.0
14.778112
2.1
2.2
17.148957
19.855030
f ′( 2 )
with
h = 0.1
24. Using the Forward Difference formula:
1
f ′ ( x0 ) ≈ { f ( x0 + h ) − f ( x0 ) }
h
1
f ′ ( 2) ≈
{ f ( 2.1) − f ( 2 ) }
0.1
1
=
{ 17.148957 − 14.778112}
0.1
= 23.708450
25. Using the 1st Three-point formula:
1
{ − 3 f ( x0 ) + 4 f ( x0 + h) − f ( x0 + 2h) }
f ′( x0 ) ≈
2h
1
[ − 3 f ( 2) + 4 f ( 2.1) − f ( 2.2)]
f ′( 2 ) ≈
2 × 0.1
1
[ − 3 × 14.778112 + 4 × 17.148957
=
0.2
− 19.855030 ]
= 22.032310
26. Using the 2nd Three-point formula:
1
{ f ( x0 + h) − f ( x0 − h) }
f ′( x0 ) ≈
2h
1
[ f ( 2.1) − f (1.9)]
f ′( 2 ) ≈
2 × 0.1
1
[ 17.148957 − 12.703199
=
0.2
= 22.228790
The exact value of
f ′( 2 ) is : 22.167168
]
27. Comparison of the results with h = 0.1
The exact value of f ′ ( 2 ) is 22.167168
Formula
f ′ ( 2)
Error
Forward Difference
23.708450
1.541282
1st Three-point
22.032310
0.134858
2nd Three-point
22.228790
0.061622
28. Second-order Derivative
h2 ( 2 )
h3 ( 3 )
f ( x + h) = f ( x ) + hf ′( x ) +
f ( x) +
f ( x) +
2
3!
2
3
h ( 2)
h ( 3)
f ( x − h) = f ( x ) − hf ′( x ) +
f ( x) −
f ( x) +
2
3!
Add these two equations.
2h 2 ( 2 )
2h4 ( 4 )
f ( x + h) + f ( x − h) = 2 f ( x ) +
f ( x) +
f ( x) +
2
4!
2h2 ( 2 )
2 h4 ( 4 )
f ( x + h) − 2 f ( x ) + f ( x − h) =
f ( x) +
f ( x) +
2
4!
29. f ( x + h) − 2 f ( x ) + f ( x − h)
2h ( 4 )
( 2)
= f ( x) +
f ( x) +
2
h
4!
2
f ( x + h ) − 2 f ( x ) + f ( x − h ) 2h 2 ( 4 )
f ( 2) ( x ) =
−
f ( x) +
2
h
4!
f
( 2)
f ( x + h) − 2 f ( x ) + f ( x − h)
( x) ≈
h2
30. NUMERICAL INTEGRATION
b
∫ f ( x )dx =
a
area under the curve f(x) between
x = a to x = b.
In many cases a mathematical expression for f(x) is
unknown and in some cases even if f(x) is known its
complex form makes it difficult to perform the integration.
33. Area of the trapezoid
The length of the two parallel sides of the trapezoid
are: f(a) and f(b)
The height is b-a
b
∫
a
b−a
[ f ( a ) + f ( b )]
f ( x )dx ≈
2
h
= [ f ( a ) + f ( b )]
2
42. Riemann Sum
The area under the curve is subdivided into n
subintervals. Each subinterval is treated as a
rectangle. The area of all subintervals are added
to determine the area under the curve.
There are several variations of Riemann sum as
applied to composite integration.
43. ∆ xIn Left Riemann
= ( b − a) / n
sum,
x1 = a the left-
x2
x3
side sample of
= afunction is
the + ∆ x
used as the
= a + 2∆the
x
height of
individual
M
rectangle.
xi = a + ( i − 1) ∆ x
∫
b
a
n
f ( x ) dx ≈ ∑ f ( xi ) ∆ x
i =1
44. ∆x = ( b − a) / n
In Right Riemann
sum, the right-side
x1 = a + ∆ x
sample of the
x2 = a is used
function+ 2∆ x as
the height of the
x3 = a + 3∆ x
individual rectangle.
M
xi = a + i∆ x
n
∫ f ( x ) dx ≈ ∑ f ( x ) ∆ x
b
a
i =1
i
45. In the ) / n
∆ x = ( b − aMidpoint
Rule, the sample at
x1 = the (middle1of ∆ x / 2 )
a + 2 ×1 − ) ( the
x2 =subinterval1is( used 2 )
a + ( 2× 2 − ) ∆x /
as the height of the
x3 = individual 1) ( ∆ x / 2 )
a + ( 2×3−
M rectangle.
xi = a + ( 2 × i − 1) ( ∆ x / 2 )
n
∫ f ( x ) dx ≈ ∑ f ( x ) ∆ x
b
a
i =1
i
46. Composite Trapezoidal Rule:
Divide the interval into n subintervals and apply
Trapezoidal Rule in each subinterval.
b
∫
a
h
f ( x )dx ≈ f ( a ) + 2∑ f ( x k ) + f (b )
2
k =1
n −1
where
b−a
h=
and x k = a + kh for k = 0, 1, 2, ... , n
n
47. π
Find
∫ sin (x)dx
0
by dividing the interval into 20 subintervals.
n = 20
b−a π
h=
=
n
20
kπ
x k = a + kh =
, k = 0, 1, 2, ....., 20
20
48. π
h
∫ sin( x )dx ≈ 2 f ( a ) + 2∑ f ( xk ) + f (b)
k =1
0
n −1
19
π
kπ
=
sin( 0 ) + 2∑ sin 20 + sin( π )
40
k =1
= 1.995886
49. Composite Simpson’s Rule:
Divide the interval into n subintervals and apply
Simpson’s Rule on each consecutive pair of subinterval.
Note that n must be even.
b
∫
a
h
f ( x )dx ≈ f ( a ) + 2
3
n/ 2
+ 4∑
k =1
( n / 2 ) −1
∑ f (x
k =1
2k
)
f ( x 2 k −1 ) + f (b )
50. where
b−a
h=
and x k = a + kh for k = 0, 1, 2, ... , n
n
π
Find
∫ sin (x)dx
0
by dividing the interval into 20 subintervals.
b−a π
n = 20
h=
=
n
20
kπ
x k = a + kh =
, k = 0, 1, 2, ....., 20
20